递归指数法 [英] Recursive Exponent Method

问题描述

public static int exponent(int baseNum) {
    int temp = baseNum *= baseNum;                

        return temp * exponent(baseNum);             
}

现在，如果我调试它，上面的方法会将n * n转换为无穷大，所以它仍然可以工作，但我需要这个递归方法在10次后停止，因为我的教练要求我们找到给定10的幂的指数。

Right now the method above does n * n into infinity if I debug it, so it still works but I need this recursive method to stop after 10 times because my instructor requires us to find the exponent given a power of 10.

该方法必须只有一个参数，这里是调用指数的一些例子：

The method must have only one parameter, here's some examples of calling exponent:

                System.out.println ("The power of 10 in " + n + " is " + 
                    exponent(n));

因此输出应为：

The power of 10 in 2 is 1024

OR

The power of 10 in 5 is 9765625

推荐答案

创建一个辅助方法来进行递归。它应该有两个参数：base和exponent。为指数调用值10，并使用（exponent-1）递归。基本情况是 exponent == 0 ，在这种情况下它应该返回1.（你也可以使用 exponent == 1 作为基本情况，在这种情况下它应该返回基数。）

Create an auxiliary method to do the recursion. It should have two arguments: the base and the exponent. Call it with a value of 10 for the exponent and have it recurse with (exponent-1). The base case is exponent == 0, in which case it should return 1. (You can also use exponent == 1 as a base case, in which case it should return the base.)

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